# Lesson video

In progress...

Hello, welcome to today's math lesson, with me, Miss Jones.

Hope you're feeling well today.

I'm feeling great and can not wait to get started.

Before we do start, let's warm up our brains.

I've got a riddle for you.

I follow you around all the time and I copy your every move, but you cannot touch me and you cannot catch me.

What am I? I'll say it one more time.

I follow you around all the time, and I copy your every move, but you cannot touch me and you cannot catch me.

It follows you around, but you can't touch it and you can't catch it.

Now our brains are warmed up.

Let's get started with today's lesson.

In today's lesson, we'll be finding solutions to problems with two variables.

Now, variables are unknown values.

We call them variables because they could be, more than one answer sometimes.

Let's look at our agenda.

We're going to be starting by finding all possibilities when we have two variables.

Then we're going to find a pair of possibilities for an equation with two unknowns.

Then we're going to do our task and our quiz.

All you'll need today is something to write with and something to write on.

Such as a pencil and a piece of paper.

If you haven't got what you need just yet, pause the video now and go and get yourself ready.

Let's begin.

We're starting off by looking at a problem with more than one unknown.

We're going to be working systematically in order to find all the possibilities.

Let's look at our problem, it's a tasty one today.

At Nardini's Ice Cream Parlour, Elvia can get two scoops, in a cone for just £150.

As long as the two scoops are different flavours.

The available flavours are vanilla, strawberry, chocolate and pistachio.

How many possible combinations are there if Elvia wants to get two scoops for £150? How might you solve and represent this problem? And when you're doing so, how can you be sure that you found all the possibilities? I want you to pause the video and have a think about how you might represent it and work through this problem.

Let's look at some people's possible strategies here.

We've got three people, who have tried to work out the problem in three different ways.

Have a little look about how they did it.

So we've got Larry.

Larry says, he hasn't really got a representation.

He's just trying to work it out.

He just says there are two different combinations.

You can have vanilla with strawberry and you can have chocolate with pistachio.

Now he hasn't used a representation to help him work this out.

I'm not sure if he's worked systematically.

Larry says, there are six different combinations.

He's used, some pictorial representations to represent the ice cream.

Using different colours, for different flavours.

Now Suzy says, there are 12 different combinations.

I've been systematic and I found all the possibilities.

But she's ended up with a different answer, to Joe, and a different answer to Larry.

So, who's right? Have a think about who here has made a mistake.

Let's look at this a little bit more closely.

So we've got Suzy, who says there are 12.

But if we look closely at Suzy's, she's got here one possibility, vanilla and strawberry.

Vanilla and chocolate, vanilla and pistachio.

Let me come to strawberry.

She's also got strawberry and vanilla.

That's exactly the same as vanilla and strawberry.

If you were at an ice cream van and you asked for vanilla and strawberry, and they gave you strawberry vanilla, it would be the same thing.

She's got some repeats, in her workings out.

So we say that Suzy has made a mistake.

Some of these have been repeated.

Now, Larry says there's only two different combinations, but he hasn't worked systematically.

He's only done vanilla and strawberry and chocolate with pistachio, and he's missed them out.

He could have had vanilla with chocolate.

Or vanilla with pistachio, not just vanilla and strawberry.

Now Joe has worked systematically and he started with vanilla and he's tried to go through what combinations he could have with vanilla.

Vanilla and strawberry, vanilla and chocolate, vanilla and pistachio, green for pistachio.

Then he's looked at the next flavour.

Strawberry with chocolate, strawberry with pistachio.

He hasn't included vanilla and strawberry again here.

So actually, Joe was correct.

I've represented it slightly different on the next page.

Instead of using the coloured circles, I've used letters.

I'll show you what I've done.

I've started with vanilla and strawberry, vanilla and chocolate, vanilla and pistachio, then I've thought about the next flavour, strawberry.

Strawberry and chocolate, strawberry and pistachio, and I thought about the next flavour, chocolate and pistachio.

You can see a represented each one with a letter.

My method is quite similar to Joe's method and we both ended up with six combinations.

Take a moment to think about what's the same and what's different.

Now, it doesn't matter which way you represent it.

You could use letters or you could use a pictorial representation.

What's important is that we read the information carefully.

For example, the scoops have to be different flavours.

So we couldn't have had vanilla and vanilla, for example.

Make sure you work systematically to try and find all the possibilities.

Otherwise, if you just look at this and guess, you might miss out some of the possibilities.

Now let's look at an equation with two variables.

How many different pairs of numbers could be represented by the variables f and g, if both are positive integers.

It means we're not going to have any negative numbers.

An integer just means it has to be a whole number.

For example, one, two or three.

Not at fraction or a decimal number such as 1.

5.

How would you solve this problem? And how could you be sure of finding all the combinations? Pause the video now and have a think about that.

Let's look at this together.

What we need to do here, is think about our unknowns f and g.

Because there are two unknowns, we can't work this out using arithmetic straight away.

We need to think about what possibilities they could be and there are multiple different options.

So what I'm going to do is I'm going to try different numbers as the value of f.

Now, we know that our whole cannot be more than 10.

So f could be one, two or three, but it couldn't be four.

Why couldn't it be four? Because f.

if f was four, we'd have three lots of f, which would make 12.

And we still need to add on another number.

Then our answer would go above 10.

So I'm going to try out f is one, f is two and f is three, and see if I can work out what the value of g would be.

Here's what I got.

If f is one, three lots of one, is three.

Which means that g needs to be seven.

If f was two, three lots of two would be six, which means that g, would have to be four.

If f was three, three lots of three would be nine, which means g would have to be one.

This is all about an alien problem on the planet Zarg.

Sounds interesting.

On Planet Zarg, there are two species of alien.

Bipods and Tripods.

Bipods are like humans, they have two legs.

Tripods have three legs.

At a party of Bipods and Tripods, there are 19 legs in total.

How many of each species, could there be? I know that I'm dealing with positive integers here.

'Cause we can't have negative numbers when we're dealing with amounts of people or amounts of legs.

And we can't really have fractions either.

Can't have half a person.

So I'm looking for possibilities using positive integers.

I want you to think about how you'd solve this problem.

What representations might you use here, and how could you be sure, you're finding all the combinations.

I'd like you to spend a few minutes having a go at this problem.

When we come back together, we're going to go through a couple of different ways that we could solve the problem.

You might have your own way of doing it.

Pause the video now, to have a go at this problem.

I'm going to show you one way that you could solve it.

We've got Joe here.

You know Joe, is very good at working systematically.

Last time he drew his ice creams. This time, he's drawn a table.

Let's have a think about how he solved this.

So he started with Bipods.

He's gone through, but what if we had one Bipod? Or two Bipods, three Bipods.

How many legs would they have? And how many legs will the Tripod have to have, to get near to our answer of 19? So he said, if a Bipod has.

if we have one Bipod, we would have two legs.

If we have two legs, I'm going to say that, let's have five Tripods, which would make 15 legs.

The closest I can get is 17 legs, and we need 19.

So, we can't have one Bipod.

Then he tried two Bipods, which gave him four legs.

And he said, if I've got four legs, I need to get 19.

I can have five Tripods to get 15.

And then he went through different amounts of Bipods, trying to find the closest he could get to 19.

By deciding how many Tripods.

He found out that, if we have two Bipods and five Tripods, we would end up with 19, or, we could have five Bipods and three Tripods still end up with 19.

Or, we could have eight Bipods, which you get to 16 legs and one Tripod, and we'd end up with 19.

He came up with three solutions.

Now, it took him quite a long time to get all this information in the table.

So although his method was very accurate and he worked systematically, made sure he found all the possibilities.

Because there's a lot to do, it could have meant that somebody might've made an error with this method.

And you also have to think about carefully, how many Tripods there are for each one.

If we made a mistake here, it's possible we could have missed out a possibility.

So I really like his method, but, I think perhaps they could be, you could use a second method to check that you're right and double check that you've found all the combinations.

What do you think about Joe's method? I'm now going to show you the way that I work this out, which is using a slightly different strategy.

What I did, was I represented the number of Bipods and the number of Tripods with letters.

Here, I've got b representing my number of Bipods.

So 2b, shows me how many legs the Bipods have all together.

And I've got t representing the amounts of Tripods.

So 3t tells me the amount of legs the Tripods have.

We know that the total amount of legs, needs to be 19.

Now, I'm going to substitute some different numbers in, to see what happens to the other value.

I'm going to focus this time on the t.

So I'm going to try out different values of t to see what b ends up and see which ones can get me a total of 19.

The reason I've chosen t is because, simply because, it's multiplying by a bigger number.

So it might make it easier later on to divide by two.

Now rather than just trying every number, I need to think about my whole, and see if I can set some boundaries.

Now I know if I'm multiplying by three and my whole cannot be more than 19, I need to think about what t could be.

Well t could be one, two, but let's think about if t could be a bigger number.

If t was five, I know that this will be 15, that's okay.

It doesn't go over 19.

Now if t was six, this would equal 18.

That would only leave me one more, to get to 19, when I've got 2b.

So t cannot be six or more.

So I'm going to try out to see what happens when t is one, two, three, four or five, and see which ones work, and get me a total of 19.

So if t, was one, that would leave me with 2b, plus three, is equal to 19.

So that means that 2b, is equal to 16.

So b, is equal, to eight.

Let's see if that works.

So eight times two is 16, added to three is 19, that works.

So t is one and b is eight is one possibility.

Now I'm going to see what happens if I put t, is equal to two.

If t is equal to two, I get 2b, added to two lots of three, which is six, is equal to 19.

So that means 2b, needs to be six less than 19.

2b needs to be 13.

Now I cannot divide 13 by two to find b.

So therefore, this one does not work.

Let's try t being equal to three, t is equal to three, 2b plus three lots of t, which is nine.

This time is equal to 19.

So I know that 2b needs to be nine less than 19.

2b, needs to be, 10.

So b, would have to be five.

That seems to work, let's test it out.

So we've got, 2b, so two lots of five is 10, plus nine, which is three lots of three, is equal to 19.

That works.

Now we're going to try t equals four.

I'm going to make a conjecture.

Last time I used an even number two, I ended up with a number here, that wasn't divisible by two.

I think that could happen again, but I'm still going to find out.

So 2b, plus 3t, which this time is 12, three lots of four.

It's equal to 19.

Which means 2b needs to be equal to seven.

Now to find out b, I can't divide, seven by two.

Seven isn't divisible, by two.

So this one isn't going to work.

And then finally, I'm going to try t equals five.

So that leaves me with 2b, add it to 15, equals 19.

So that means 2b, is equal to four, and b is equal to two.

That one seems to work as well.

So, t equals five and b equals two, let's try it out.

2b, would get me to four, added to 15, just three lots of five, is equal to 19.

That works, so these are my three possibilities, t equals one, b equals eight, t equals three, b equals five, and t equals five and b equals two.

So I solve that using letters to represent the number of Bipods and Tripods.

And Joe, used a table.

It doesn't matter which method you use, as long as you're being systematic and checking you found all the possibilities.

What's also helpful, is to think of those boundaries.

What couldn't my whole be? Is there a maximum or minimum perhaps that I should be, working with? Rather than, if I would have just tried t as 20, it wouldn't have worked because we would have got too big of a whole.

To finish off this, let's explore section.

I just want you to spend a moment reflecting, how did you work systematically? How did, I and some of the examples I showed you, work systematically to find the answer? Were any of the strategies similar or different? Were any of the questions similar or different? What did you notice? Take a moment to reflect.

Here we've got a different kind of problem.

This time, we've got some shapes.

A pentagon and a triangle representing our unknowns.

This represents a number and this represents a number.

And again, we've got a problem here where there could be multiple possibilities.

Now, if I add another piece of information, I can solve this problem with one possibility.

How has this changed our problem? Here, we could have had multiple possibilities, but now we know, that two pentagons equal 12.

We can find out what one pentagon is, and then we can use that, to help us find out the second variable, in the other equation.

For example, if two pentagons equals 12, one pentagon has to equal six.

Because six plus six, equals 12.

Now I know that one pentagon equals six.

I've got a new equation.

Six plus triangle plus triangle, equals 12.

Now if I subtract, the hexagon here, I need to balance it by subtracting six on this, part of my equation as well.

So I could say that two triangles must be equal to six.

Then we can work out that one triangle, is equal to three.

Sometimes an extra piece of information can help me work out my second variable.

Let's have a look at this other problem.

Here we've got Andy looking at some tile patterns made from two size tiles.

He knows the width of the patterns and.

but not the width of the tiles.

There's two unknowns, he doesn't know the width of the green tile or the white tile, but he knows the overall width.

Now if we just had this pattern, we wouldn't be able to work out the exact value of two variables.

Just the different possibilities.

But because we have this second piece of information, we can use this to help us, because this pattern is only, one white tile more than this pattern.

And we know that this total is two more than this.

So therefore, our white tile needs to have a value of two.

Another way I could represent this is using letters.

I've got one, two, three, four green tiles.

I'm just going to use the letter g, plus, three white tiles is equal to 26.

Down here I've got four green tiles again, but this time I've got four white tiles and it's equal to 28.

I can see that my second equation only has one more white tile.

Which means my white tile needs to be the difference between my two answers, which is two.

Once I know that the white tile is two, I can substitute that in.

So three white tiles or 3w, must equal six.

So we can write a new equation here.

I'm just going to use the top one to help me this time.

And I know that 4g, add six, is equal to 26.

So that means 4g, is equal to 20.

Six less than 26.

If I know that 4g is equal to 20, I can divide by four, to find the value of my green tile.

So my green tile, needs to be five.

Let's see if that makes 26? Five, I'm just going to count in fives and then twos, because that's a little bit easier.

Five, 10, 15, 20, 22, 24, 26.

Now that works.

We could have used our bar models to help us, or we can use our letters.

But sometimes it's nice to use a combination of the two to check that you've got the right answer.

For your main task, I'm going to show you, several problems. I want you to think about which ones might have one solution and which ones have more than one solution.

And of course, have a go at working out the different solutions or, the sole solution to each one.

So I'd like you to pause the video now, and go and complete your main task.

Let's have a look at some of the solutions together.

For this one we had two equations.

We have two variables, and we have two bits of information about them.

We have to work out the values of x and y.

Now if I'm looking at just one equation at a time, I can see that there are different possible values.

I'm just looking at this first equation and I found that x could be one, y could be five and that would work, x could be two, and y could be three and that would work, x could be three and y be one, and that would work.

And I've just written some notes to explain why.

Now there's multiple possibilities for this equation, but when we look at our second bit of information here, that might help us to think about which of these possibilities might work for both.

So I'm going to now look at all of my x and y values and think about, do any of them add together to equal five? And we can see that one plus five equals six so not that one.

Two plus three does add together to equal five.

Three plus one doesn't.

So actually, this is the only one that satisfies both pieces of information.

With my second piece of information, I can work out the exact value, of x and y, with x being equal to two and y being equal to three.

Let's have a look at the next one.

We know that a and b are both positive integers.

Record, the correct pairs of value for a and b.

So I'll show you some of my solutions here.

I found out that if a was four, b could be equal to five.

If a was six, b would be equal to eight.

Shown my workings up if you want to have a look.

If a was eight, b would equal 11.

Now I'm noticing a pattern that my odd numbers weren't working, while my even numbers worked.

I heard there's another pattern, that actually, each time I'm going up, you know in even numbers are two, but my b number is also going up by an equal amount.

It's going up three each time.

So now I can make a conjecture that this pattern might continue.

If a was 10, b would be equal to 14.

However, I need to remember, that a and b have to be one digit numbers.

So actually, b could not equal 11 and b could not equal 14, and we can't keep this going.

So the only two solutions to this, with this additional piece of information, are this one, a equals four and b equals five, and this one, a equals six and b equal two.

Interestingly, you could have also put a equal to two.

But that would have ended up with b, also equaling two.

Which could be a possibility.

But usually when we have two different letters, they represent two different numbers.

So that's an interesting one.

Let's have a look at this problem.

We've got here, two oranges, which weigh the same.

Have the same mass.

And that also is equal to, 'cause you can see this is a balance, one orange, plus 100 grammes.

Now I've represented this in a bar model this time.

Got my two apples, which is the same as an orange, plus 100.

Now this piece of information is crucial.

We know that the mass of the orange is 30 grammes greater than one apple.

So we know that this part of our bar, has to be equal to 30.

Once we know that we can think about working out the mass of our apples and oranges.

Here, I know that, two apples which I've called a, is the same as orange plus 100.

But, I know an additional piece of information.

I know my orange, is the same as an apple plus 30.

Or a plus 30.

So instead of putting orange into the equation, I'm going to put in, a plus 30, which is the same as, the orange.

Now this way, I only have one unknown.

It's repeated in the equation.

But hopefully, I can use some arithmetic to find out what the value of a is.

A, one a has to equal 30, plus 100, which is equal to 130.

So if an apple is equal to 130, we know that our orange is 130, plus another 30.

Here.

So our orange has to be 160.

Our apples are 130.

Finally, this one.

I've got two bits of information.

So I'm hoping there's just one solution to this one.

Let's have a look.

We can look at these star sort of shapes here with, more points, or blue star.

We can notice three of them that total is 102.

So we can divide by three, to find out what one of those blue shapes is.

One of those blue shapes is equal to 34.

Once we know that, we can look at this one and you can find it out as an equation if you want, and two stars or two of something plus 34, need to total to 100.

So let's look at this, and we've got, let's take away that 34.

So then we just got two stars, is equal to 66.

That one star, if we divide by two, should be equal to 33.

So the answer should be 34 for this one, 33 for this one.

That brings us to the end of the lesson.